我正在尝试创建一个对非二叉整数树的值求和的函数。
-- datastructures.hs
data Tree a = Empty | Node a [Tree a] deriving (Eq, Show)
myNums :: (Num a) => Tree a
myNums = Node 1 [
Node 2 [
Node 4 [Empty], Node 5 [Empty]
],
Node 3 [
Node 6 [Empty], Node 7 [Empty], Node 8 [Empty]
]
]
addNums :: (Num a) => Tree a -> a
addNums Empty = 0
addNums (Node n [Empty]) = n
addNums (Node n (x:xs)) = n + (addNums x) + (addNums xs)
理想情况下，我想要addNums myNums to be 36，但这会产生错误：
datastructures.hs:20:54:
Couldn't match expected type ‘Tree a’ with actual type ‘[Tree a]’
Relevant bindings include
xs :: [Tree a] (bound at datastructures.hs:20:20)
x :: Tree a (bound at datastructures.hs:20:18)
n :: a (bound at datastructures.hs:20:15)
addNums :: Tree a -> a (bound at datastructures.hs:18:1)
In the first argument of ‘addNums’, namely ‘xs’
In the second argument of ‘(+)’, namely ‘(addNums xs)’
我该如何应对这个问题，最佳做法是什么？
编辑：最佳实践似乎省略Empty共！我忘记了[]是类型的有效实例[Tree a]。所以实现这个的最好方法是：
data Tree a = Node a [Tree a] deriving (Eq, Show)
addNums :: (Num a) => Tree a -> a
addNums (Node n []) = n
addNums (Node n (x:xs)) = n + (addNums x) + addNums (Node 0 xs)
只需推导即可Foldable并利用现有的sum:
{-# LANGUAGE DeriveFoldable #-}
data Tree a = Empty | Node a [Tree a] deriving (Eq, Show, Foldable)
myNums :: (Num a) => Tree a
myNums = ...
main = print $ sum myNums
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